A 0.027 kg golf ball moving at 16.8 m/s crashes through the window of a house in 4.6 × 10−4 s. After the crash, the ball contin- ues in the same direction with a speed of 12.0 m/s.

Assuming the force exerted on the ball by the window was constant, what was the mag- nitude of this force?
Answer in units of N

To find the magnitude of the force exerted on the golf ball by the window, we can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (mΔv) over time (Δt).

The formula is: F = (mass of the golf ball) × ((final velocity of the golf ball) - (initial velocity of the golf ball)) / (time taken)

Given:
- Mass of the golf ball (m) = 0.027 kg
- Initial velocity of the golf ball (u) = 16.8 m/s
- Final velocity of the golf ball (v) = 12.0 m/s
- Time taken (Δt) = 4.6 × 10^(-4) s

Substituting these values into the formula, the calculation becomes:

F = (0.027 kg) × ((12.0 m/s) - (16.8 m/s)) / (4.6 × 10^(-4) s)

To simplify the calculation, first, subtract the two velocities and then divide it by the time.

F = (0.027 kg) × (-4.8 m/s) / (4.6 × 10^(-4) s)

Next, multiply the mass and velocity to get the numerator of the fraction.

F = -0.1296 kg·m/s / (4.6 × 10^(-4) s)

Finally, divide the numerator by the time to obtain the magnitude of the force.

F = -0.1296 kg·m/s ÷ (4.6 × 10^(-4) s)

Using the appropriate conversion factors, simplify the equation:

F = -0.1296 kg·m/s ÷ (4.6 × 10^(-4) s)
= -0.1296 kg·m/s × 10^4 s / (4.6)
= -1296 kg·m/s ÷ 4.6
= -282.6 N

The magnitude of the force exerted on the golf ball by the window is approximately 282.6 N.